The line L is tangent to a circle, center at (2.5), at point (-1,1). Suppose line L is passing through (-5,4). Determine the equation of the line L.
If the required line L passes through the points (-1,1) and (-5,4), then it's equation is
(x-(-5)) / (-1-(-5)) = (y-4) / (1-4)
(x+5)/4 = (y-4)/(-3)
-3(x+5) = 4(y-4)
-3x - 15 = 4y - 16
4y + 3x = 1.
Let's check whether this line is tangent to a circle center at (2,5). If so, it will be normal to the line that passes through (2,5) and (-1,1):
(x-(-1)) / (2-(-1)) = (y-1) / (5-1)
(x+1)/3 = (y-1)/4
4(x+1) = 3(y-1)
4x + 4 = 3y - 3
3y - 4x = 7.
Here is the the scalar product of the direction vectors:
(4,3)*(3,-4) = 4*3 + 3*(-4) = 0.
We see that it is tangent to a circle. So, required line equation is 4y + 3x = 1.
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